Question

create a grouped frequency distribution using the following data. separate the data into eight classes. (include the class limit, class boundaries, tally, and frequency).

1. the follow data represents the high temperature for each day in june.

88 79 64 88 85 70 110 80 75 88
100 65 80 60 105 99 90 56 102 77
71 78 55 70 70 75 95 85 85 72
bonus: add the relative frequency and class midpoint to the table.

Explanation:

Step1: Find the range

The range \(R\) is calculated as \(R=\text{Max}-\text{Min}\). Here, \(\text{Max} = 110\) and \(\text{Min}=55\), so \(R=110 - 55=55\).

Step2: Calculate the class width

The class width \(w\) is given by \(w=\frac{R}{k}\), where \(k = 8\) (number of classes). \(w=\frac{55}{8}\approx7\).

Step3: Determine class limits

Start with the minimum value.

• Class 1: \(55 - 61\)
• Class 2: \(62 - 68\)
• Class 3: \(69 - 75\)
• Class 4: \(76 - 82\)
• Class 5: \(83 - 89\)
• Class 6: \(90 - 96\)
• Class 7: \(97 - 103\)
• Class 8: \(104 - 110\)

Step4: Calculate class boundaries

For a class \(a - b\), the lower boundary is \(a-0.5\) and the upper boundary is \(b + 0.5\).

• Class 1: \(54.5 - 61.5\)
• Class 2: \(61.5 - 68.5\)
• Class 3: \(68.5 - 75.5\)
• Class 4: \(75.5 - 82.5\)
• Class 5: \(82.5 - 89.5\)
• Class 6: \(89.5 - 96.5\)
• Class 7: \(96.5 - 103.5\)
• Class 8: \(103.5 - 110.5\)

Step5: Tally and frequency

• Class 1 (\(55 - 61\)): Tally \(\vert\vert\) (values \(55,56,60\)), Frequency \(3\)
• Class 2 (\(62 - 68\)): Tally \(\vert\vert\) (values \(64,65\)), Frequency \(2\)
• Class 3 (\(69 - 75\)): Tally \(\vert\vert\vert\vert\vert\) (values \(70,70,70,71,72,75,75\)), Frequency \(7\)
• Class 4 (\(76 - 82\)): Tally \(\vert\vert\vert\vert\) (values \(77,78,79,80,80\)), Frequency \(5\)
• Class 5 (\(83 - 89\)): Tally \(\vert\vert\vert\vert\vert\vert\) (values \(85,85,85,85,88,88,88,88\)), Frequency \(8\)
• Class 6 (\(90 - 96\)): Tally \(\vert\vert\) (values \(90,95,99\)), Frequency \(3\)
• Class 7 (\(97 - 103\)): Tally \(\vert\vert\) (values \(100,102\)), Frequency \(2\)
• Class 8 (\(104 - 110\)): Tally \(\vert\) (value \(105,110\)), Frequency \(2\)

Step6: Calculate class mid - point

The class mid - point \(x_m=\frac{\text{Lower limit}+\text{Upper limit}}{2}\)

• Class 1: \(\frac{55 + 61}{2}=58\)
• Class 2: \(\frac{62+68}{2}=65\)
• Class 3: \(\frac{69 + 75}{2}=72\)
• Class 4: \(\frac{76+82}{2}=79\)
• Class 5: \(\frac{83 + 89}{2}=86\)
• Class 6: \(\frac{90+96}{2}=93\)
• Class 7: \(\frac{97+103}{2}=100\)
• Class 8: \(\frac{104 + 110}{2}=107\)

Step7: Calculate relative frequency

The relative frequency \(rf=\frac{\text{Frequency}}{n}\), where \(n=3+2+7+5+8+3+2+2 = 32\)

• Class 1: \(rf=\frac{3}{32}\approx0.09\)
• Class 2: \(rf=\frac{2}{32}=0.06\)
• Class 3: \(rf=\frac{7}{32}\approx0.22\)
• Class 4: \(rf=\frac{5}{32}\approx0.16\)
• Class 5: \(rf=\frac{8}{32}=0.25\)
• Class 6: \(rf=\frac{3}{32}\approx0.09\)
• Class 7: \(rf=\frac{2}{32}=0.06\)
• Class 8: \(rf=\frac{2}{32}=0.06\)

Answer:

Class Limit	Class Boundaries	Tally	Frequency	Class Mid - point	Relative Frequency
\(62 - 68\)	\(61.5 - 68.5\)	\(\vert\vert\)	\(2\)	\(65\)	\(0.06\)
\(69 - 75\)	\(68.5 - 75.5\)	\(\vert\vert\vert\vert\vert\)	\(7\)	\(72\)	\(0.22\)
\(76 - 82\)	\(75.5 - 82.5\)	\(\vert\vert\vert\vert\)	\(5\)	\(79\)	\(0.16\)
\(83 - 89\)	\(82.5 - 89.5\)	\(\vert\vert\vert\vert\vert\vert\)	\(8\)	\(86\)	\(0.25\)
\(90 - 96\)	\(89.5 - 96.5\)	\(\vert\vert\)	\(3\)	\(93\)	\(0.09\)
\(97 - 103\)	\(96.5 - 103.5\)	\(\vert\vert\)	\(2\)	\(100\)	\(0.06\)
\(104 - 110\)	\(103.5 - 110.5\)	\(\vert\)	\(2\)	\(107\)	\(0.06\)